The crystal of NaCl (ccp type) contains 7.5272 × 10^23 unit cells, then what would be the weight of that crystal?

To determine the weight of a crystal of sodium chloride (NaCl) that contains 7.5272 × 10^23 unit cells, we first need to understand a few key concepts about unit cells, molar mass, and the relationship between these quantities.

Understanding the Basics

A unit cell is the smallest repeating unit in a crystal lattice that reflects the overall symmetry and structure of the entire crystal. In the case of NaCl, it crystallizes in a face-centered cubic (fcc) structure, which is also referred to as a cubic close-packed (ccp) structure.

Molar Mass of NaCl

The molar mass of sodium chloride can be calculated by adding the atomic masses of sodium (Na) and chlorine (Cl). Sodium has an atomic mass of approximately 22.99 g/mol, and chlorine has an atomic mass of about 35.45 g/mol. Therefore, the molar mass of NaCl is:

• Na: 22.99 g/mol
• Cl: 35.45 g/mol
• Total Molar Mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

Calculating the Number of Moles

In a face-centered cubic structure, each unit cell contains 4 formula units of NaCl. To find out how many moles of NaCl are present in 7.5272 × 10^23 unit cells, we first calculate the total number of formula units:

• Total Formula Units = Number of Unit Cells × 4
• Total Formula Units = 7.5272 × 10^23 × 4 = 3.01088 × 10^24

Next, we convert the number of formula units to moles using Avogadro's number, which is approximately 6.022 × 10^23 units/mol:

• Number of Moles = Total Formula Units / Avogadro's Number
• Number of Moles = 3.01088 × 10^24 / 6.022 × 10^23 ≈ 5.000 moles

Calculating the Weight of the Crystal

Now that we have the number of moles, we can find the weight of the crystal by multiplying the number of moles by the molar mass of NaCl:

• Weight = Number of Moles × Molar Mass
• Weight = 5.000 moles × 58.44 g/mol = 292.2 g

Final Result

Thus, the weight of the crystal of NaCl containing 7.5272 × 10^23 unit cells is approximately 292.2 grams. This calculation illustrates the relationship between unit cells, moles, and mass in a crystalline structure, providing a clear example of how to approach similar problems in crystallography and materials science.